Question

7sin²A+ 4cos²A=4 then find the value of tanA​

Answer 1

Answer:

➡7sin²A+4cos²A=4

➡7sin2A+4-4sin²A=4

➡3sin²A=0

➡sinA=0

➡A=0°

therefore, tanA=tan0°=0

Answer 2

Answer:

7sin²A+4cos²A=4

7sin²A+4-4sin²A=4

3sin²A. =0

sinA=0

A. =0°

so,

tanA=tan0°=0

hope it helps!!!